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$ p \geq q \geq 1$, then $L^p_{[a,b]} \subset L^q_{[a,b]}$

I want to prove it with Holder inequality for integrals. However I am not sure how to proceed. This is what I did:

Let $f \in L^p_{[a,b]}$. Then I can write: $$ \int_a^b \vert f(x) \vert^p dx < \infty $$

$$ \int_a^b \vert f(x) \vert^p dx = \int_a^b \vert f(x) \vert^{p-q+q} dx $$ $$ \int_a^b \vert f(x) \vert^{p+q} \vert f(x) \vert^{-q}dx $$ But from here I do not know how to proceed. I am not even sure this is the right way to approach the problem.

EDIT: As suggested in the comments I tried to prove that $\frac{1}{p} = \frac{1}{q} + \frac{1}{r}$ and Holder inequality gives: $\Vert fg \Vert _p \leq \Vert f \Vert_q \Vert g \Vert_r$

By holder inequality and the equality that I have written before I can write: $$ \Vert fg \Vert_{\frac{1}{p}} \leq \Vert f \Vert_{\frac{1}{q}} \Vert g \Vert_{\frac{1}{p}} $$

Then the inequality follows. However I am struggling to see how I can write $\Vert f \Vert_q$ less than something that converges.

qcc101
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  • Essentially this: https://math.stackexchange.com/q/66029/587192 –  Nov 20 '18 at 21:41
  • I saw that, but I want to prove it using strictly things from functional analysis. – qcc101 Nov 21 '18 at 06:53
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    @qcc101 Does Holder inequality counts as functional analysis? – Jacky Chong Nov 21 '18 at 06:55
  • Yes, but the link by @user587192 has a flavor of measure theory and I want something strictly analytical if that makes sense. – qcc101 Nov 21 '18 at 06:56
  • Do you mean you are not comfortable with abstract measure theory? Because you won't be able to avoid measure theory completely - that's just how the $L^p$ spaces are defined. And the arguments given the answers are exactly the same as for the linked question. – MaoWao Nov 21 '18 at 13:22

2 Answers2

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Hint: Prove the following: If $p, q, r\geq 1$ and \begin{align} \frac{1}{p}=\frac{1}{q}+\frac{1}{r} \end{align} then \begin{align} \|fg\|_p \leq \|f\|_q\|g\|_r. \end{align}

Additional Hint: The proof of the above hint uses Holder inequality. Once you are done proving the hint set $g=\ldots$. I will let you figure this out.

Jacky Chong
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  • I do not see how from that inequality I have $\Vert f \Vert_q$ less than something that converges. Can you shed some light? – qcc101 Nov 21 '18 at 06:52
  • At first I tried putting $g = g/f$ but I don't think it works – qcc101 Nov 21 '18 at 09:16
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Using Holder's inequality one has $$ \|f\|_q^q=\int_a^b |f(x)|^q\cdot 1\ dx\le \big\Vert|f|^q\big\Vert_{p/q}\|1\|_{p/(p-q)}=\Vert f\Vert_p^q(b-a)^{(p-q)/p} $$